Question

How many terms of the A.P:27,23,19,15....have the sum 95? ​

Answer 1

Answer:

5th

watch explanation

Answer 2

Answer: 5th term.

Step-by-step explanation:

The sum of nth term of Arithmetic Progression

Sn = n/2{ (2a + (n - 1)d , where a = first term, d is the common difference of the progression. From the question, Sn = 95, a = 27

But to find d, always subtract the second term from the first term.

So from here

d = 23 - 27

= -4

Now , solving the question , we have

95 = n/2 { (2(27) + (n - 1) x (-4)

95 = n/2 { 54 -4n + 4 }

Multiply both side by 2 to make it a linear expression

190 = n { 58 - 4n }

Open the bracket by multiplying by n .

190 = 58n - 4n², now, rearrange the expression to give a quadratic expression/ equation.

4n² - 58n + 190 = 0

Using any method to solve

4n² - 38n - 20n + 190 = 0

By factor

2n(2n - 19) - 10(2n - 19) = 0

(2n - 19) is common

(2n - 19)(2n - 10) = 0

solving for n, the root

n = 19/2 or 5

Therefore, n could = 19/2 or 5.

Now check the roots or results to see the one the gives ,95.

Using 5

5/2( 2 x27 + 4(-4)}

5/2 (54 -16)

5/2 x 38

= 190/2

= 95.

So 19/2 is term extraneous.

Therefore, the 5th term of that progression will give 95.